Even though you were correct that 0.999... is equal to one, after reading the discussion on Wikipedia some more I think the method of proof that you used here (with an infinite geometric series) only proves the case for the limit like I said.
The problem I have with the proof using infinite geometric series like you used is illustrated in the second picture below. It doesn't say that r raised to the nth power equals 0 as n tends to infinity it just says that r raised to the nth power approaches 0. That means that the sum of the series in Example 1 (and in your proof) only approaches a / (1 - r).
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Some more information from my real analysis book (see 16.3 f) seems to indicate that x to the power of n only equals 0 when x is equal to 0 and only approaches 0, as n tends to infinity, when the absolute value of x is less than one.
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The information on Wikipedia is also a bit confusing:
Quote:
...the definition that still dominates today: the sum of a series is defined to be the limit of the sequence of its partial sums. A corresponding proof of the theorem explicitly computes that sequence; it can be found in any proof-based introduction to calculus or analysis.[5]
A sequence (x0, x1, x2, ...) has a limit x if the distance |x- xn| becomes arbitrarily small as n increases. The statement that 0.999... = 1 can itself be interpreted and proven as a limit:
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